of 1 98
!
Copyright © 2022 by Ian Beardsley
ISBN: 978-1-387-56187-2
of 2 98
Contents
The Basis………………………………………………………………..3
1.0 Introduction………………………………………………………11
2.0 Giordano’s Relationship…………………………………….15
3.0 The Principle of Least Action……………………………..24
4.0 Proton-Seconds………………………………………………..30
5.0 The Natural Second…………………………………………..36
6.0 Rigorous Formulation of Proton Seconds……………40
7.0 The Theory……………………………………………………….47
8.0 Orbital Velocity………………………………………………..63
9.0 Logos………………………………………………………………74
10.0 Conclusion………………………………………………77
11.0 Summary…………………………………………………82
12.0 Graphic Equations………………………………………85
Matter can be explained as a construct of space and time at
the fundamental level of elementary particles, the elements
of the periodic table follow from that construct. We
demonstrate a recurrent six-fold symmetry of Nature and
formulate logos, her language in abstract terms.
of 3 98
The Basis The transformation that
rotates counter-clockwise where is given
by the standard matrix
Equation 1
We suggest there is an aspect of Nature founded on
six-fold symmetry, the example of which we are
interested in here is The Periodic Table of the
Elements, because it has 18 groups which we can
define by carbon, C. This because we have the
following scenario:
Equations 2
And, we pull out the 2 and the 3 and write (Fig. 1)
, ,
Where , such that
T : ℝ
2
→ ℝ
2
T(
x ) = A
x
A =
(
cos(θ) −sin(θ)
sin(θ) cosθ
)
3 + 3 + 3 = 9
2 + 2 + 2 = 6
3 ⋅ 6 = 18
2 ⋅ 9 = 18
2 ⋅ 3 = 6
2cos
π
4
= 2
2cos
π
5
=
5 + 1
2
2cos
π
6
= 3
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
of 4 98
which is given by
In general
Equations 3 ,
, ,
And these can be mapped by the matrix A onto a linear
vector
space (Fig. 2)
=
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
f (n) = 2cos
π
n
n = 4,5,6
π
4
= 45
∘
π /5 = 36
∘
π
6
= 30
∘
A =
(
2cos(θ ) −2sin(θ )
2sin(θ ) 2cosθ
)
A =
(
2cos(30
∘
) −2sin(30
∘
)
2sin(30
∘
) 2cos(30
∘
)
)
A
e
1
=
3 −1
1 3
(
1
0
)
= ( 3,1)
Fig. 1 Dividing line in
golden mean,
of 5 98
=
A
e
2
=
3 −1
1 3
(
0
1
)
= (−1, 3)
A =
(
2cos(36
∘
) −2sin(36
∘
)
2sin(36
∘
) 2cos(36
∘
)
)
Φ −
1
2
(5 − 5)
1
2
(5 − 5) Φ
A
e
1
= Φ,
1
2
(5 − 5)
A
e
2
=
1
2
(5 + 5), Φ
Fig. 2!
of 6 98
=
Our is based on the square (Fig. 3)
is the line . The reflection through
is given by:
Equation 4
And our is the equilateral triangle:
A =
(
2cos(45
∘
) −2sin(45
∘
)
2sin(45
∘
) 2cos(45
∘
)
)
2 − 2
2 2
A
e
1
= 2, 2
A
e
2
= − 2, 2
2cos
π
4
π
4
= 45
∘
x
2
= x
1
x
2
= x
1
A =
(
0 1
1 0
)
2cos
π
6
Fig. 3
of 7 98
To transform the square into the equilateral triangle
we expand the square of base with the matrix
Equation 5
And we see becomes and we have added half
the square to itself. (Fig. 4)
Or, better we can use the contraction
We draw in the diagonal of the the half-square and
reassemble the two half-triangles into an equilateral
e
1
A =
(
3/2 0
0 1
)(
1
0
)
=
3
2
,1
e
1
3/2
A =
(
1/2 0
0 1
)(
1
0
)
=
1
2
Fig. 4
of 8 98
triangle (Fig. 8). To get we take the half square and
draw in the circle of radius 1/2. (Fig. 5) We have
Thus we see the periodic table is 18 groups (Fig. 6).
Carbon is in group 14. We have 18-14=4 valence
electrons. Hydrogen is neither a metal or a non-metal
but ionizes like a metal by losing one electron
becoming and carbon being means it needs 4
positive ions to be neutral meaning it combines with 4
hydrogens to each C, or with two hydrogens to a C and
a C in long chains (hydrocarbons) which form the
Skeltons of organic compounds in life chemistry (Fig.
7) .
Φ
1
2
2
+ 1
2
=
1
4
+
4
4
=
5
2
5
2
+
1
2
=
5 + 1
2
= Φ
H
+
C
4−
Fig. 5
of 9 98
Fig. 7
Fig. 8
Fig. 6
of 10 98
That is:
is phenomenal because It allows multiplication
between degrees and seconds to output our
fundamental ratios ( ). We see in the
following wave:
, ,
Where t=1 second is carbon yielding:
And, t=6 seconds is hydrogen yielding:
1
6protons
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.00second
2, 3, . . .
A = A
0
cos(θt)
A
0
= 1
θ = 30
∘
,60
∘
,45
∘
A(60
∘
) = cos(60
∘
⋅ 1s) = 0.5
A(30
∘
) = cos(30
∘
⋅ 1s) = 3/2
A(45
∘
) = cos(45
∘
⋅ 1s) = 2/2
A(60
∘
) = cos(60
∘
⋅ 6s) = 1
A(30
∘
) = cos(30
∘
⋅ 6s) = − 1
A(45
∘
) = cos(45
∘
⋅ 6s) = 0
of 11 98
1.0 Introduction Gravity is a property of space measured
by the universal constant of gravity, G:
Equation 1.1
Matter, or inertia, which measures matter’s ability to resist
a force is for each particle (protons and neutrons) we will
suggest given by:
Equation 1.2.
Which describes mass per meter over time, which is:
Equation 1.3:
It must be adjusted by the fine structure constant . It is
my guess the factor should be which is 18,769.:
Equation 1.4.
Here we are suggesting that the proton and neutron are the
3-dimensional cross-sections of a hypersphere. Thus we
consider the surface area of a proton, :
Equation 1.5.
We take the square root to get meters:
Equation 1.6.
We multiply that with the value we have in equation 1.4:
G = 6.674E − 11N
m
2
s
2
= 6.674E − 11
m
3
s
2
⋅ kg
h
Gc
=
kg ⋅ s
m
1.82E − 16
kg ⋅ s
m
α
1
α
2
(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s /m)
S
p
S
p
= 4π r
2
p
= 8.72E − 30m
2
S
p
= 2.953E − 15m
of 12 98
Equation 1.7.
We find that the mass of a proton realizes if we divide
this by six seconds:
Equation 1.8.
That is hydrogen. We see that the element carbon
manifests if we divide instead by 1 second:
Equation 1.9.
Carbon (C) is the core element of life chemistry and it
combines with hydrogen (H) to make the skeletons of
organic matter, the so-called hydrocarbons:
Equation 1.10.
Equation 1.11.
If we divide 1E-26kgs by something greater than 6 seconds
we get fractional protons. The rest of the elements in the
periodic table occur for dividing by something less that 1-
second. It seems the duration of a second is natural. If it is,
since it was formed by a calendar based on reconciling the
periods of the moon and the sun in the earth sky, it should
be in the Earth-moon orbital mechanics. I find it is, that:
That is, the earth day (86,400 seconds) times the kinetic
energy of the moon to the kinetic energy of the earth is
about 1 second (about 1.2 seconds).
(2.953E − 15m)(3.416E − 12
kg ⋅ s
m
) = 1.009E − 26kg ⋅ s
m
p
m
p
=
1E − 26
6secon d s
= 1.67E − 27kg
m
c
=
1E − 26kg ⋅ s
1secon d
= 1E − 26kg = 6pr oton s = 6m
p
m
p
=
1E − 26kg ⋅ s
6secon d s
m
c
=
1E − 26kg ⋅ s
1secon d
K E
moon
K E
earth
(Ear th Da y) ≈ 1secon d
of 13 98
All of the computations thus far are shown on the next
page…
of 14 98
of 15 98
2.0 Giordano’s Relationship We can find our one
second in terms of the atom that predicts the mass of
a proton. But I find we need to develop a constant I
call k, which I derived from Giordano’s relationship.
We now formulate what I call Giordano’s Relationship:
Warren Giordano writes in his paper The Fine Structure
Constant And The Gravitational Constant: Keys To The
Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes
analyzing Einstein’s geometric to kinematic equations,
along with an observation that multiplying Planck’s
constant ‘ ’ by ‘ ’, where ‘ ’ is the Fine Structure
Constant, and multiplying by yielded Newton’s
gravitational constant numerically, but neglecting any
units.
Let’s do that
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js
And it works, G is:
G=6.67408E-11 N(m2/kg2)
Let us reformulate this as:
Equation 2.1
Where
and H=1 gram/atom
Because for hydrogen 1 proton is molar mass 1 gram, for
carbon 6 protons is 6 grams and so on for 6E23 atoms per
gram. Thus,…
h
1 + α
α
10
23
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
N
A
= Avaga dr o′ s − Num ber = 6.02E 23atom s /gra m
of 16 98
Since grams and atom cancel we can work in grams even
though our equations are in kilograms. Let us not write H,
since formally it is grams per mole of hydrogen but write
We have:
Or,…
Equation 2.2
Where
Equation 2.3
Let us say we were to consider Any Element say carbon
. Then in general
Equation 2.4
We have
and
Because there are six grams of protons in carbon which has
6 protons and 6 neutrons and a molar mass of 12. We have
N
A
H = 6.02E 23
atom s
gram
⋅
1gram
atom
= 6.02E 23
ℍ = 1
gram
atom
h
(1 + α)
G
⋅ N
A
ℍ = 6.0003kg
2
⋅
s
m
h(1 + α)N
A
ℍ = 6Gx
x = 1.00kg
2
s
m
𝔼
ℂ
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6gram s
6proton s
N
A
=
6(6E 23proton s)
6gram s
of 17 98
12-6=6 grams of protons in the 12 grams of protons and
neutrons. Thus
And it follows that
We see in general since the atomic number Z is the number
of protons in an atom that in general this holds for all
elements because
And,
Therefore we always have:
Equation 2.5
This works nicely because we formulated molar mass
nicely; we said element one (hydrogen) which is one
proton and one electron has one gram for a mole of atoms.
Historically this was done because we chose carbon
(element six) to have 12 grams per mole, and determined
what the mole was such that it would hold. The reason this
works is that hydrogen is one proton and has no neutrons,
but carbon has twelve neutrons but since hydrogen doesn’t
have any neutrons, and the neutron has the same mass as
the proton, and our theory makes use only of protons (in
this instance of its formulation) equation 2.3
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gram s
𝔼 =
Z ⋅ gram s
Z ⋅ proton s
N
A
⋅ 𝔼 = 6E 23
x = 1.00kg
2
s
m
of 18 98
Comes out to have x equal to 1.00 (nearly) even. It is at this
moment that we point out, because it is important, that in
equation 2.5
is not molar mass, and that is a variable determined
by ; it is the number of a mole of atoms multiplied by the
number of protons in . The reason we point this out,
though it may already be clear, is we wish to find the
physical theory behind it. That is we need to find the
physical explanation for equation 2.4
It is the integer 6 to 3 ten thousandths. Which classifies it
as interesting because since it is in kilograms, seconds, and
meters, it may mean these units of measurement have
some kind of a meaning. We can in fact write it:
We know that
The fine structure constant squared is the ratio of the
potential energy of an electron in the first circular orbit to
the energy given by the mass of an electron in the Bohr
model times the speed of light squared. To begin our
search for the meaning of equation 1.4 we convert x, the
factor of 1.00 to astronomical units, years, and
solar masses, as these are connected to the orbit of earth as
it relates to the sun. We have:
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
𝔼
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.000kg
2
⋅
s
m
α
2
=
U
e
m
e
c
2
kg
2
⋅
s
m
of 19 98
=
We can now write
Eq 2.6.
This unit of AU/year is very interesting. It is not
, which would be the Earth’s orbital velocity,
but is a velocity given by the earth orbital radius to its
orbital period, which is quantum mechanical in nature. It
relates to earth as as a state, as we have with atoms, a
number. We multiply both sides by and we have earth
velocity on the left and the units stay the same on the right.
But what we will do is return to the form in kg-m-s and
leave it as an equation but put in the Earth mean orbital
velocity which is 29.79km/s (Zombeck, Martin V. 1982).
We get:
Eq. 2.7
This brings up an interesting question: while we have
masses characteristic of the microcosmos like protons, and
masses characteristic of the macrocosmos, like the
minimum mass for a star to become a neutron star as
opposed to a white dwarf after she novas (The
Chandrasekhar limit) which is 1.44 solar masses, we do not
have a characteristic mass of the intermediary world where
we exist, a truck weighs several tons and tennis ball maybe
around a hundred grams. To find that mass let us take the
geometric mean between the mass of a proton and the
kg
2
1
⋅
s
m
⋅
(1.98847E 30)
2
M
2
⊙
kg
2
1.4959787E11m
AU
⋅
year
3.154E 7s
1.8754341E64
M
2
⊙
⋅ year
AU
h
(1 + α)
G
⋅ N
A
𝔼
AU
year
= 8.2172E 32M
⊙
2π AU/year
4π
2
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
of 20 98
mass of 1.44 solar masses. We could take the average, or
the harmonic mean, but the geometric mean is the
squaring of the proportions, it is the side of a square with
the area equal to the area of the rectangle with these
proportions as its sides. We have:
We multiply this by 1.44 to get 2.8634E30kg. The mass of a
proton is . We have the
intermediary mass is:
Eq. 2.8
All we really need to do now is divide 2.7 by 2.8 and we get
an even number that is the six of our six-fold symmetry.
Eq. 2.9
The six of our six-fold symmetry.
We have something very interesting here. We have
This is:
Equation 2.10
Where k is a constant, given
Equation 2.11
We can take the velocity of earth as being 30,000 m/s by
rounding it. We have
M
⊙
= 1.98847E 30kg
m
p
= 1.67262E − 27kg
m
i
= (2.8634E 30)(1.67262E − 27) = 69.205kg
1
m
i
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 6.1092 ≈ 6
1
69.205kg
6kg
2
s
m
v
e
= 6
k v
e
= 6
k =
1
800
s
m
of 21 98
30,000
800
= 37
1
2
37.5 = 6.123734357
1
6
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
of 22 98
Using , we write
Eq. 2.12
The primordial element from which all others were made.
We can explicitly write the constant k:
Equation 6.13
It was the Indian physicist Chandrasekhar who found the
limit in mass for which a white dwarf will not have its
gravity overcome the degeneracy pressure and collapse.
The non-relativistic equation is:
k v
e
= 6
1
α
2
m
p
h 4π r
2
p
Gck v
e
= 1proton ⋅ secon d
(K . E . Moon)(Ear th Da y)
(K . E . E ar th)
≈ 1secon d
1
α
2
m
p
h 4π r
2
p
Gck v
e
(K . E . E ar th)
(K . E . Moon)(Ear th Da y)
= 1proton
1
α
2
m
p
h 4π r
2
p
Gc
(K . E . E ar th)
(K . E . Moon)(Ear th Da y)
= 6proton s
1
α
2
m
p
h 4π r
2
p
Gck v
e
(K . E . E ar th)
(K . E . Moon)(Ear th Da y)
= ℍ
ℍ = hydrogen
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼
of 23 98
Equation 2.14
Let us approximate 0.77 with 3/4. Since we have our
constant
And
Then
Equation 2.15
Since our constant k in terms the Chandrasekhar
limit is
Equation 2.16
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
k =
1
m
2
i
h
1 + α
G
⋅ N
A
𝔼
m
i
= Mm
p
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2 π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
of 24 98
3.0 The Principle of Least Action We consider two
paths, one with velocity c in one medium, the other with
velocity v in another. In order to go from one point two
another over two paths, the refraction is such that the sine
of the angle of incidence equals the sine of the angle of
refraction.
We have the two paths are travelled in a time t:
This is the mysterious Nature of reality, the path of least
time is taken. This falls under the general heading The
Principle of Least Action, attributed to firstly the French
Natural Philosopher and mathematician Louis Maupertuis
of the early eighteenth century. I say mysterious, because
as it is said in physics, for something to know the path of
least action and take it, it is as if it has explored first all
paths between A and B to know which one would be the
path of least action. Everything in physics comes to this
principle. It is called a principle, not a theory, law, or rule.
Yet it seems to be the way Nature behaves, and it is
mysterious. Richard Feynman applied it to quantum
mechanics, probably because the mysterious Planck’s
constant that governs quantum mechanics is in Joule-
seconds, energy over time, and this is the terms in which
action is formulated mathematically.
In our scenario here we regard matter, the proton in
particular, as the cross-section of a hypersphere. Our two
mediums are hyperspace and space and the least action
principle applies in the same mathematical form. This is
abstract cosmology, that really is the underlying
mathematics is common to all systems, that in effect they
are manifestations of one another.
We have
sinα
sinβ
=
v
c
t = r
p
c + v
c v
of 25 98
We make the approximation
So that
We are saying t=6 seconds is the proton, and r is the radius
of a proton r=0.833E-15m. Thus
The radius of a hydrogen atom is .
And we have our one second as a natural constant with
respect to the atom. We see it occurs at t/2 which is at half
the radius of a hydrogen atom. We want to deal with that
size or around it; our velocity comes from it and it can be
thought of as a proton drift velocity akin to the electron
drift velocity in a wire that gives rise to an electrical
t
r
c v − v = c
t
r
c v = c
v =
r
t
v =
0.833E − 15m
6s
= 1.389E − 16m /s
R
h
= 1.2E − 10m
t =
1.2E − 10m
1.389E − 16m /s
= 863,930.8855s
t
2
= 431,965.4428s
t
k
= (773.5m /s)(431,965.4428s) = 334,125,270m
334,125,270m
c
=
334,125,270m
299,792,459m /s
= 1.11452193secon d s
of 26 98
current. We find if we derive the equation that represents
these computations, we have
Where,…
, ,
Radius of hydrogen atom
Remember our constant k equation 4.16 (Don’t forget to
divide by two somewhere):
v =
r
t
t = 6s =
1
α
2
m
p
h 4π r
2
p
Gc
1
t
= α
2
m
p
Gc
h 4π r
2
p
v =
r
p
t
= α
2
m
p
Gc
h 4π
R
H
= 1.2E − 10m
t =
R
h
v
= R
h
⋅
1
α
2
m
p
h 4π
Gc
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
t
ck
=
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12second s
of 27 98
Since we have the equation of the radius of a proton is
given by, by evaluating it at one second which is carbon:
And 1 second in terms of the atom is given by
Then the equation for the radius of a proton is:
Equation 3.1.
And since
Then,…
Equation 3.2.
Let’s verify equation 3.1:
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
= 1.12secon d s
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
= 1.12secon d s
1
α
2
m
p
h 4π r
2
p
Gc
≈ 6 ⋅
K Eof Moon
K Eof Ear th
⋅ Ear th Da y = (6)1.2secon ds
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
=
K Eof Moon
K Eof Ear th
⋅ Ear th Day
r
p
=
9
8
2
1
1.67E − 27
(6.626E − 34)(299,792,459)
(6.674E − 11)4π
3
⋅
1.2E − 10
6.02E 23
= 0.93f m
of 28 98
Remember that in equation 3.1, which is
That, must remain coupled. is determined by the
number of protons in .
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
N
A
𝔼
N
A
𝔼
of 29 98
of 30 98
4.0 Proton-Seconds We show carbon, the core
element of life is six-fold symmetric with hydrogen in
terms of the natural constants that characterize space,
time, and matter:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
We find one second gives six protons which is carbon:
Equation 4.1
We find six seconds gives 1 proton is hydrogen:
Equation 4.2
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G : 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792,459m /s
α : 1/137
1
α
2
m
p
h4π r
2
p
Gc
= 6proton ⋅ seconds = carbon(C )
1
α
2
m
p
h4πr
2
p
Gc
= 1proton ⋅ 6secon ds = hydrogen(H )
of 31 98
For time t greater than 6 seconds we have fractional
protons. For t<6 we the have other elements.
Is proton-seconds. Divide by time we have a number
of protons because it is a mass divided by the mass of
a proton. But these masses can be considered to
cancel and leave pure number. We make a program
that looks for close to whole number solutions so we
can create a table of values for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in
equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
1
α
2
m
p
h4π r
2
p
Gc
of 32 98
A very interesting thing here is looking at the values
generated by the program, the smallest integer value 1
second produces 6 protons (carbon) and the largest
integer value 6 seconds produces one proton
(hydrogen). Beyond six seconds you have fractional
protons, and the rest of the elements heavier than
carbon are formed by fractional seconds. These are
the hydrocarbons the backbones of biological
chemistry. Here is the code for the program:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27,
h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to
increment?: ");
scanf("%f", &increment);
printf("How many values would you like to
calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/
(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
of 33 98
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart
\n", protons[i], t-increment, decpart);
}}}}
We need an interpretation of equation 2.1, which was
Matter is that which has inertia. This means it resists
change in position with a force applied to it. The more
of it, the more it resists a force. We understand this
from experience, but what is matter that it has
inertia?
In this analogy we are suggesting a proton is a three
dimensional bubble embedded in a two dimensional
plane. As such there has to be a normal vector holding
the higher dimensional sphere in a lower dimensional
space. Thus if we apply a force to to the cross-section
of the sphere in the plane there should be a force
countering it proportional to the normal holding it in
a lower dimensional universe. It is actually a 4-
dimensional hypersphere whose cross-section is a
sphere. This counter force would be experienced as
inertia. (Fig. 1.1)
Since plank’s constant h is a measure of energy over
time where space and time are concerned it must play
a role. Of course the radius of a proton plays a role
1
α
2
m
p
h4π r
2
p
Gc
= 6proton ⋅ seconds = carbon(C )
of 34 98
since squared and multiplied by it is the surface
area of our proton embedded in space. The
gravitational constant is force produced per kilogram
over a distance, thus it is a measure of how the
surrounding space has an effect on the proton giving
it inertia. The speed of light c has to play a role
because it is the velocity at which events are separated
through time. The mass of a proton has to play a role
because it is a measurement of inertia itself. And alas
the fine structure constant describes the degree to
which these factors have an effect. We see the inertia
then in equation 6 is six protons over 1 second, by
dimensional analysis.
4π
of 35 98
We see the radius of a proton is given by carbon (1
second):
Equation 4.3.
The experimental radius of a proton is:
Equation 4.4.
The fine structure constant squared is the ratio of the
potential energy of an electron in the first circular
orbit to the energy given by the mass of an electron in
the Bohr model times the speed of light squared:
Equation 4.5.
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
r
p
= 0.833
±
0.014 f m
α
2
=
U
e
m
e
c
2
Fig 1.1
of 36 98
5.0 The Natural Second In that we get one second
for carbon and 6 seconds for hydrogen very nearly
even, that is
Eq. 5.1
It is suggested that the second is a natural unit. If it is,
since it comes from designing a calendar that
reconciles the phases of the moon with the Earth year
(12 moons per year, approximately) it is suggested the
unit of a second should be in the Earth-Moon-Sun
orbital mechanics. The translational kinetic energy of
the moon and earth are:
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
1
6
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.004996352seconds
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
2π (3.85E8m) = 2.419E 9m
E =
1
2
mv
2
of 37 98
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.
It turns out:
Eq . 5.2
Where the Lunar Month can be as much as 31 days
and is based on the lunar orbital period (27.32 days).
We have
Eq. 5.3
Essentially we have formed a Planck constant, h, for
the moon by multiplying its kinetic energy over the
time for the period of its orbit:
If we let the lunar month cancel with moon’s orbital
period, we have:
2π (1.496E11m) = 9.399E11m
1second ≈
(KEof Moon)(Lun arOrbitalPeriod )
Luna rMonth
EarthDay
(KEof Ear th)
31days
(1Ear th Day)
= 31 ≈ π
3
= 31.006
h = (3.67E 28J )(2.36E6s) = 8.6612E 34J ⋅ s
8.6612E 34J ⋅ s
KEof Ear th
= 32.696seconds
of 38 98
Eq. 5.4
Since
is units of mass divided by we can let it cancel with
, the mass of a proton, and write:
Eq. 5.5
That is:
is phenomenal because It allows multiplication
between degrees and seconds to output our
fundamental ratios ( ). We see in the
following wave:
, ,
Where t=1 second is carbon yielding:
1
α
2
m
p
h4πr
2
p
Gc
≈ 6protons ⋅
KEof Moon
KEof Ear th
⋅ Ear th Day
1
1sec
⋅
1
α
2
m
p
h4π r
2
p
Gc
m
p
m
p
1
α
2
m
p
h4π r
2
p
Gc
≈ 6 ⋅
KEof Moon
KEof Ear th
⋅ Ear th Day = (6)1.2secon ds
1
6protons
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.00second
2, 3, . . .
A = A
0
cos(θt)
A
0
= 1
θ = 30
∘
,60
∘
,45
∘
of 39 98
And, t=6 seconds is hydrogen yielding:
In so far as
relates carbon=1second to the Earth-Moon-Sun
orbital mechanics and to the radius of a proton
through six-fold symmetry:
We have the following around which all our structure
is based
,
, ,
A(60
∘
) = cos(60
∘
⋅ 1s) = 0.5
A(30
∘
) = cos(30
∘
⋅ 1s) = 3/2
A(45
∘
) = cos(45
∘
⋅ 1s) = 2/2
A(60
∘
) = cos(60
∘
⋅ 6s) = 1
A(30
∘
) = cos(30
∘
⋅ 6s) = − 1
A(45
∘
) = cos(45
∘
⋅ 6s) = 0
1
α
2
m
p
h4π r
2
p
Gc
≈ 6 ⋅
KEof Moon
KEof Ear th
⋅ Ear th Day ≈ (6)1second
1
α
2
m
p
h4π r
2
p
Gc
= 6
f (n) = 2cos
π
n
n = 2,3,4,5.6.,...
f (2) = 2
f (5) = Φ = 1/ϕ
f (6) = 3
of 40 98
6.0 Rigorous Formulation of Proton Seconds
We can actually formulate this differently than we have.
We had
But if t1 is not necessarily 1 second, and t6 is not
necessarily six seconds, but rather t1 and t2 are lower and
upper limits in an integral, then we have:
Equation 6.1
This Equation is the generalized equation we can use for
solving problems.
Essentially we can rigorously formulate the notion of
proton-seconds by considering
Equation 6.2
Is protons-seconds squared where current density is
and ( can also be ). We say
Equation 6.3
Keeping in mind q is not charge (coulombs) but a number
of charges times seconds, here a number of protons. It is
1
t
1
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton s
1
t
6
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
∫
t
qdt = t
2
∬
S
ρ(x, y, z) d x d y
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρd V
of 41 98
Equation 6.4
Dividing Equation 6.2 through by by t:
Equation 6.5
Which is proton-seconds. Dividing through by t again:
Equation 6.6
We see that if where and then
J is I/m2 (current per square meter) is analogous to
amperes per per square meter which are coulombs per
second through a surface. Thus we are looking at a number
of protons per second through a surface. Thus we write:
Is carbon where 0.5 seconds is magnesium (Mg) from the
values of time corresponding to protons in the output from
our program and 1.0 seconds is carbon (C). We see we have
the following theorem:
Equation 6.7
ℕ =
1
α
2
m
p
h 4π r
2
p
Gc
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z) d x d y
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
2
= proton s
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
of 42 98
So as an example,…
Is fluorine (F). Divide by xy with x=y=1 and we have
current density. And multiply by 1 second which is carbon
and we have protons per square meter.
1
α
2
m
p
h 4πr
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
pr otons
secon d
J(x, y, z) = (0,0, J ) = − J
k
d
S = d x d y
k
J ⋅ d
S = (0,0, J ) ⋅ (0,0, d x d y) = − Jd x d y
of 43 98
We are now equip to do computations in
proton-seconds. We use equation 7.6 from two
to three, the smallest prime numbers that
multiply to make six-fold symmetry in our
hexagonal proton that we found described its
radius (My feeling is we introduce the factor of
2 because carbon is 6 protons +6 neutrons
and 2 times 6 is twelve):
2
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
2
= protons
of 44 98
But we already know this is the radius of a proton in fm, it
is as if fm is a natural length, as well as the second we have
been using. 2 seconds in our program is 3 protons (Lithium
Li) and 3 seconds is two protons is helium, He:
By what value would you like to increment?: 0.01
How many values would you like to calculate for t in
equation 1 (no more than 100?): 100
86.1425 protons 0.070000 seconds 0.142517 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
43.0713 protons 0.140000 seconds 0.071259 decpart
40.1998 protons 0.150000 seconds 0.199841 decpart
30.1499 protons 0.200000 seconds 0.149876 decpart
26.2173 protons 0.230000 seconds 0.217283 decpart
25.1249 protons 0.240000 seconds 0.124893 decpart
24.1199 protons 0.250000 seconds 0.119900 decpart
23.1922 protons 0.260000 seconds 0.192213 decpart
20.0999 protons 0.300000 seconds 0.099920 decpart
17.2285 protons 0.350000 seconds 0.228504 decpart
15.0749 protons 0.400000 seconds 0.074944 decpart
14.0232 protons 0.430000 seconds 0.023204 decpart
13.1086 protons 0.460000 seconds 0.108647 decpart
12.0600 protons 0.500000 seconds 0.059957 decpart
11.1666 protons 0.540000 seconds 0.166627 decpart
10.2203 protons 0.590000 seconds 0.220302 decpart
10.0500 protons 0.600000 seconds 0.049964 decpart
9.1363 protons 0.660000 seconds 0.136332 decpart
8.1486 protons 0.740000 seconds 0.148621 decpart
8.0400 protons 0.750000 seconds 0.039971 decpart
7.1785 protons 0.839999 seconds 0.178546 decpart
7.0941 protons 0.849999 seconds 0.094093 decpart
7.0116 protons 0.859999 seconds 0.011603 decpart
6.2165 protons 0.969999 seconds 0.216474 decpart
6.1530 protons 0.979999 seconds 0.153040 decpart
6.0909 protons 0.989999 seconds 0.090889 decpart
2
α
2
m
p
h4πr
2
p
Gc
∫
3
2
dt
t
3
= 0.833protons/second
of 45 98
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in
equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
Thus we write
Equation 6.8.
Thus we suggest the radius of a proton of a proton
(0.833protons) is the proton radius . Now we
integrate from Aluminum is 13 protons=0.372 seconds to
phosphorus is 15 protons = 0.396 seconds which is to
integrate across silicon:
By what value would you like to increment?: 0.006
How many values would you like to calculate for t in equation 1
(no more than 100?): 100
251.2490 protons 0.024000 seconds 0.248978 decpart
59.1174 protons 0.102000 seconds 0.117416 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
40.1998 protons 0.150000 seconds 0.199844 decpart
37.2221 protons 0.162000 seconds 0.222080 decpart
2
α
2
m
p
h 4π r
2
p
Gc
∫
t
He
t
Li
dt
t
3
= 0.833proton s /secon d = r
pro ton
r
proton
of 46 98
27.1621 protons 0.222000 seconds 0.162058 decpart
25.1249 protons 0.240000 seconds 0.124907 decpart
20.0999 protons 0.300000 seconds 0.099916 decpart
17.0338 protons 0.354000 seconds 0.033823 decpart
16.2096 protons 0.372000 seconds 0.209604 decpart
15.2272 protons 0.396000 seconds 0.227202 decpart
14.1549 protons 0.426000 seconds 0.154862 decpart
13.2236 protons 0.456000 seconds 0.223620 decpart
13.0519 protons 0.462000 seconds 0.051885 decpart
12.1084 protons 0.498000 seconds 0.108375 decpart
11.1666 protons 0.540000 seconds 0.166615 decpart
11.0439 protons 0.546000 seconds 0.043906 decpart
10.1515 protons 0.594000 seconds 0.151471 decpart
Equation 6.9.
Silicon can be doped with phosphorus to make negative (n-
type) silicon that semi-conducts thus enabling the
construction of logic circuits that you can use to make
computing machines. But this must be joined with positive
(p-type) silicon which usually uses boron, but boron is in
the same group as aluminum, just above it. This results in a
theory for AI elements as mathematical constructs, that we
will go into now.
2
α
2
m
p
h4πr
2
p
Gc
∫
t
Al
t
P
dt
t
3
= − 6(6.377 − 7.22) = 5protons/secon d = boron
of 47 98
7.0 The Theory
Above we see the artificial intelligence (AI) elements pulled
out of the periodic table of the elements. As you see we can
make a 3 by 3 matrix of them and an AI periodic table.
Silicon and germanium are in group 14 meaning they have
4 valence electrons and want 4 for more to attain noble gas
electron configuration. If we dope Si with B from group 13
it gets three of the four electrons and thus has a deficiency
becoming positive type silicon and thus conducts. If we
dope the Si with P from group 15 it has an extra electron
and thus conducts as well. If we join the two types of
silicon we have a semiconductor for making diodes and
transistors from which we can make logic circuits for AI.
As you can see doping agents As and Ga are on either side
of Ge, and doping agent P is to the right of Si but doping
agent B is not directly to the left, aluminum Al is. This
becomes important. I call (As-Ga) the differential across
Ge, and (P-Al) the differential across Si and call Al a
dummy in the differential because boron B is actually used
to make positive type silicon.
That the AI elements make a three by three matrix they can
be organized with the letter E with subscripts that tell what
element it is and it properties, I have done this:
of 48 98
Thus E24 is in the second row and has 4 valence electrons
making it silicon (Si), E14 is in the first row and has 4
valence electrons making it carbon (C). I believe that the AI
elements can be organized in a 3 by 3 matrix makes them
pivotal to structure in the Universe because we live in three
dimensional space so the mechanics of the realm we
experience are described by such a matrix, for example the
cross product. Hence this paper where I show AI and
biological life are mathematical constructs and described
in terms of one another.
We see, if we include the two biological elements in the
matrix (E14) and and (E15) which are carbon and nitrogen
respectively, there is every reason to proceed with this
paper if the idea is to show not only are the AI elements
and biological elements mathematical constructs, they are
described in terms of one another. We see this because the
first row is ( B, C, N) and these happen to be the only
elements that are not core AI elements in the matrix,
except boron (B) which is out of place, and aluminum (Al)
as we will see if a dummy representative makes for a
mathematical construct, the harmonic mean. Which means
we have proved our case because the first row if we take the
cross product between the second and third rows are, its
respective unit vectors for the components meaning they
describe them!
The Computation
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
A = (Al, Si, P)
B = (G a, Ge, As)
of 49 98
And silicon (Si) is at the center of our AI periodic table of
the elements. We see the biological elements C and N being
the unit vectors are multiplied by the AI elements, meaning
they describe them. But we have to ask; Why does the first
row have boron in it which is not a core biological element,
but is a core AI element? The answer is that boron is the
one AI element that is out of place, that is, aluminum is in
its place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to
make diodes and transistors, silicon (Si) and germanium
(Ge) doped with boron (B) and phosphorus (P) or gallium
(Ga) and arsenic (As) have an asymmetry due to boron.
Silicon and germanium are in group 14 like carbon (C) and
as such have 4 valence electrons. Thus to have positive type
A ×
B =
B
C
N
Al Si P
Ga Ge A s
= (Si ⋅ A s − P ⋅ G e)
B + (P ⋅ G a − Al ⋅ As)
C + (Al ⋅ G e − Si ⋅ G a)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g/mol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g/mol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsinθ = (50)(126)sin8
∘
= 877.79
877.79 = 29.6g/m ol ≈ Si = 28.09g/m ol
of 50 98
silicon and germanium, they need doping agents from
group 13 (three valence electrons) like boron and gallium,
and to have negative type silicon and germanium they need
doping agents from group 15 like phosphorus and arsenic.
But where gallium and arsenic are in the same period as
germanium, boron is in a different period than silicon
(period 2) while phosphorus is not (period 3). Thus
aluminum (Al) is in boron’s place. This results in an
interesting equation.
Equation 7.1.
The differential across germanium crossed with silicon
plus the differential across silicon crossed with germanium
normalized by the product between silicon and germanium
is equal to the boron divided by the average between the
germanium and the silicon. The equation has nearly 100%
accuracy (note: using an older value for Ge here, it is now
72.64 but that makes the equation have a higher accuracy):
Due to an asymmetry in the periodic table of the
elements due to boron we have the harmonic mean
between the semiconductor elements (by molar
mass):
Equation 7.2
This is Stokes Theorem if we approximate the harmonic
mean with the arithmetic mean:
Si(A s − G a) + Ge(P − Al )
SiGe
=
2B
Ge + Si
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(A s − G a) +
Ge
B
(P − Al ) =
2SiGe
Si + G e
of 51 98
We can make this into two integrals:
Equation 7.3.
Equation 7.4
If in the equation (The accurate harmonic mean
form):
Equation 7.5
We make the approximation
Equation 7.6
Then the Stokes form of the equation becomes
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d xd y ≈
1
Ge − Si
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Si
B
(As − Ga)dyd z ≈
1
3
1
(Ge − Si)
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Ge
B
(P − Al )d x dz ≈
2
3
1
(Ge − Si)
∫
Ge
Si
yd y
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
of 52 98
Equation 7.7
Thus we see for this approximation there are two
integrals as well:
Equation 7.8
Equation 7.9
For which the respective paths are
One of the double integrals on the left is evaluated in
moles per grams, the other grams per mole (0 to 1
moles per gram and 0 to 1 grams per mole).
By making the approximation
In
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
dyd z =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − Ga)dyd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )d ydz =
2
3
∫
Ge
Si
dz
y
1
=
1
3
B
SiGa
ln(z)
y
2
=
2
3
B
Si Al
ln(z)
2SiGe
Si + Ge
≈ Ge − Si
of 53 98
We have
Equation 7.10
is the differential across Si,
is the differential across Ge and
is the vertical differential.
Which is Ampere’s Circuit Law
We see if written
Which is interesting because it is semiconductor
elements by molar mass which are used to make
circuits.
We say (Phi) is given by
and
And
=1.618
Si( As − Ga)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ Ge
ΔSi
ΔS
= B
ΔSi = P − Al
ΔGe = As − Ga
ΔS = Ge − Si
∇ ×
B = μ
0
J + μ
0
ϵ
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
Φ
a = b + c
a
b
=
b
c
Φ = a /b
of 54 98
=0.618
(phi) the golden ratio conjugate. We also find
Equation 7.11
Thus since
And we have
Equation 7.12
We see and are both and c is in the Si (silicon)
field wave, but for E and B fields c is the speed of light.
ϕ = b/a
ϕ
(ϕ)ΔGe + (Φ)ΔSi = B
∇ ×
B = μ
J + μϵ
0
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
ΔG e =
ΔS
Si
B −
G e
Si
ΔSi
(
∇
2
−
1
c
2
∂
2
∂t
)
E = 0
(
∇
2
−
1
c
2
∂
2
∂t
)
B = 0
c =
1
ϵ
0
μ
≈ ϕ
μ
ϵ
0
Φ
ϕ
ϵ
0
= 8.854E − 12F ⋅ m
−1
μ = 1.256E − 6H /m
of 55 98
To find the Si wave our differentials are
It is amazing how accurately we can fit these differentials
with an exponential equation for the upward increase. The
equation is
This is the halfwave:
G e
Si
= μ ϵ
0
ΔS
Si
= μ
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Si = 0
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
G e = 0
ΔC = N − B = 14.01 − 10.81 = 3.2
ΔSi = P − Al = 30.97 − 26.98 = 3.99
ΔGe = As − Ga = 74.92 − 69.72 = 5.2
ΔSn = Sb − In = 121.75 − 114.82 = 6.93
ΔPb = Bi − T l = 208.98 − 204.38 = 4.6
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
of 56 98
of 57 98
Building Another Matrix
We pull these AI elements out of
the periodic table of the elements
to make an AI periodic table:
We now notice we can make a 3 by 3 matrix of it, which
lends itself to to the curl of a vector field by including
biological elements carbon C (above Si):
=
=
Which results in Stokes theorem:
Where
i
j
k
∂
∂x
∂
∂ y
∂
∂z
( − C ⋅ P) y (Si ⋅ G a)z (Ge ⋅ As)y
(Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.92) − (28.09)(69.72)
]
i +
[
(12.01)(30.97)
]
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a = ex p
(
1
Ge − Si
∫
Ge
Si
ln(x)d x
)
∇ ×
u = (Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
of 58 98
We then able to write this with product notation
While we have the AI BioMatrix
Which we use to formulate a similar
equation. We can form another 3X3
matrix we will call the electronics
matrix:
d
a =
(
zd yd z
i + yd yd z
k
)
u = ( − C ⋅ P)y
i + (Si ⋅ G e)z
j + (G a ⋅ As)y
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a =
n
n
∏
i=1
x
i
of 59 98
We can remove the 5th root sign in the above equation by
noticing
=(28.085)(72.64)(12.085)(107.8682)(196.9657)=
Where we have substituted carbon (C=12.01) the core
biological element for copper (Cu).
But since we have:
We take the ratio and have
Almost exactly 3 which is the ratio of the perimeter of
regular hexagon to its diameter used to estimate pi in
ancient times by inscribing it in a circle:
5
∏
i=1
x
i
= Si ⋅ G e ⋅ Cu ⋅ Ag ⋅ Au
523,818, 646.5
g
5
m ol
5
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u) ⋅ d
a = 170,535, 359.662(g /m ol )
5
523,818, 646.5
170,535, 359.662
= 3.0716
of 60 98
Perimeter=6
Diameter=2
6/2=3
Thus we have the following equation…
Showing The Calculation using the most accurate data
possible…
Ge=72.64
As=74.9216
Si=28.085
Ga=69.723
C=12.011
P=30.97376200
=
=
π = 3.141 . . .
π
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u) ⋅ d
a =
5
∏
i=1
x
i
(Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.9216) − (28.085)(69.723)
]
i +
[
(12.011)(30.97376200)
]
k
3,484 . 134569
(
g
m ol
)
2
i + 372.025855
(
g
m ol
)
2
k
∫
Ge
Si
∫
Ge
Si
(
3,484 . 134569
(
g
m ol
)
2
i + 372.025855
(
g
m ol
)
2
k
)
⋅
(
zd ydz
i + yd ydz
k
)
∫
Ge
Si
∫
Ge
Si
(
3,484 . 134569
(
g
m ol
)
2
⋅ z d z d y + 372.025855
(
g
m ol
)
2
⋅ yd z d y
)
of 61 98
=154,082,837.980+16,452,521.6822=
=
(28.085)(72.64)(12.085)(107.8682)(196.9657)=
Where we have substituted carbon C=12.01 for copper Cu.
We use Cu, Ag, Au because they are the middle column of
our electronics matrix, they are the finest conductors used
for electrical wire. We use C, Si, Ge because they are the
middle column of our AI Biomatrix. Si and Ge are the
primary semiconductor elements used in transistor
technology (Artificial Intelligence) and C is the core
element of biological life. We have
Perimeter/Diameter of regular hexagon = 3.00
∫
Ge
Si
3,484 . 134569
(
(72.64 − 28.085)
2
2
)
d y +
∫
Ge
Si
372.025855y ⋅ (72.64 − 28.085)d y
3458261.42924
(
g
m ol
)
4
(72.64 − 28.085) + 16575.6119695
(
g
m ol
)
3
(
(72.64 − 28.085)
2
2
)
170,535, 359.662
(
g
m ol
)
5
5
∏
i=1
x
i
= Si ⋅ G e ⋅ C ⋅ A g ⋅ Au
523,818, 646.5
g
5
m ol
5
523,818, 646.5
170,535, 359.662
= 3.0716
π = 3.141 . . .
3.141 + 3.00
2
= 3.0705
of 62 98
The same value as our 3.0716 if taken at two places after
the decimal.
of 63 98
8.0 Orbital Velocity Earth rotates through 15 degrees in
one hour:
The distance then that the earth equatorial surface rotates
through is where r is the earth radius and theta is in
radians.
Now we look at how many degrees through which the earth
rotates in 1 minute:
For seconds…
Now let’s look at the same for the distance the earth moves
around the Sun in an hour, a minute, and a second as
well…
360
24
= 15
s = r θ
s = (6,378.14)(0.2618ra d ) = 1669.8k m /hr
24(60) = 1440min
360
1440
= 0.25
∘
= 0.0043633r a d
s = (6,378.14)(0.00436) = 27.83k m /min
24(60)(60) = 86,400s
360
∘
86,400
= 0.004167
∘
= 0.00007272r a d
s = (6,378,14)(0.00007272) = 0.464k m /s
(365.25)(24) = 8766hr /yr
360
8766
= 0.041
∘
= 0.00071678r a d
of 64 98
And finally…
We have the following table:
As we can see I am in good agreement with Martin
Zombek, Handbook of Space Astronomy and Astrophysics
which provides data. Notice 27.83 km/min in earth
rotation is approximately the 29.786 km/sec in earth orbit.
That is a valuable clue. Now let us consider
Earth: 1 AU=149,598,000km (average earth-sun
separation).
(149,598,000)(0.000716768) = 107,227k m /hr
(365.25)(24)(60) = 525960
360
525960
= = 0.000684463
∘
= 0.000011946r a d
s = (149,598,000)(0.000011946) = 1,787.1k m /min
(365.25)(24)(60)(60) = 31557600
360
31557600
= 0.000011408
∘
= 0.000000199r a d
(149,598,000)(0.000000199) = 29.786k m /sec
of 65 98
The earth rotates through about one degree a day.
Earth Orbit:
Earth Rotation: r=6,378.14km
Moon: sidereal month =27.83 days, r=405,400 km
The points to be made in this exploration
360
365.25
= 0.9856
deg
d a y
≈ 1
deg
d a y
360
∘
= 2π ra di a n s = 6.283
ra d
year
0.9856
∘
= 0.017202424
ra d
d a y
(149,598,000)(0.017202424) = 2,573,448.201
k m
d a y
360
∘
1d a y
= 6.283
ra d
d a y
s = r θ = (6,378.14)(6.283) = 40,075.0
k m
d a y
(27.83)(24) = 667.92
h ours
m onth
360
667.92
= 0.529
∘
= 0.0094
ra d
hr
(405,400k m)(0.0094ra d /hr) = 3,810.76k m /hr
(3,810.75k m /hr)(hr /60min) = 63.5k m /min = 1.0585k m /s
of 66 98
1. So the earth goes through 1 degree a day and the moon
1 kilometer per second.
2. Earth rotates through 27.83 km/min at its equatorial
surface and orbits through 29.786km/s around the sun.
These two are close to the same.
3. The synodic month is 29.53 days approximately equals
the 29.786 km/s the Earth moves around the sun. The
sidereal month is 27.83 days is the 27.83 km/min
through which the Earth rotates at its surface.
With that information we have this mystery of sexagesimal
in the earth-moon-sun orbital parameters, solved. We see
we can make the following equation in table form:
Consider minute times day:
Thus,..
The circle is divided into 360 units, each unit (each degree)
is the distance the earth moves around the Sun in a day,
where a day is one turn of the earth on its its axis, and as
such there are 360 such turns in the time it takes the earth
to go around the sun approximately (365.25 days). We
have:
1 astronomical unit (AU) is the distance of the earth from
the sun on the average, and is always close to that because
its orbit is approximately circular. We have
orbit rotat ion orbit m oon
29.786k m minute 1d a y k ilom eter
secon d 27.83k m degree secon d
=
mi n ⋅ d a y
deg
⋅
k m
s
2
(min)(d a y) = 60(24 ⋅ 60 ⋅ 60) = 864,000sec
2
min ⋅ d a y
deg
⋅
k m
s
2
=
86,400s
2
deg
⋅
k m
s
2
= 86,400 ⋅
k m
deg
86,400 ⋅
k m
deg
⋅ 360
∘
= 311,040,000k m
of 67 98
This is approximately the diameter of the Earth orbit. We
define our variables:
Earth orbits:
Earth rotates:
Earth orbits:
Moon orbits:
Earth completes a 360 degree orbit yields:
Where on the right it is in radians and is the
radius of the Earth’s orbit. We have
This is (0.00618)360=2.225
0.00618 is
311,040,000k m
149,598,000k m /AU
= 2.079AU
v
e
= 29.786
k m
s
ω
e
=
27.83k m
min
=
27.83k m
min
⋅
min
60sec
= 0.4638
k m
sec
θ
e
=
1d a y
deg
=
(24)(60)(60)
deg
= 14,400
sec
deg
v
m
=
1k m
sec
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅
360
∘
1AU
= 2π
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅ r
e
r
e
= 1AU
(29.786)(14,400)(1k m /s)
(0.4638)(149,598,000)
360 = 2.225AU
ϕ
100
of 68 98
Where is the inverse of the golden ratio.
.
We have:
Equation 8.1
Our base ten counting is defined
is defined
, such that
which is given by
Thus since the diameter of the Earth orbit is
Then its radius is
Since we measure time with the Earth orbital period and
that period is given by Kepler as
Then
ϕ
ϕ = 0.618 =
1
Φ
ϕ
100
360
∘
= 2AU
10
0
= 1,10
1
= 1,10
2
= 100,...
ϕ
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
ϕ
100
360
∘
= 2AU
1
2
⋅
ϕ
100
360
∘
= 1AU
T
2
= a
3
of 69 98
Equation 8.2
Is approximately one year. In this section we set out two
show the historical development of the second by dividing
up the motions of the Earth, Moon, and the apparent
motion of the Sun into units of 24, and 60 result in the
solar system’s size as based around the Earth orbit giving
us
Where this duration of a second is in the atomic world in
carbon (six protons) the core element of life which we
found was:
And that the duration of a second was as we showed in the
orbital mechanics of the Earth and moon as such
Since we have established a connection between the
microcosmos and the macrocosmos we would do well to
introduce the units of AU (astronomical unit), year, solar
masses. Thus we want to know the universal gravitational
constant in these units:
T =
3
2
10 ⋅
ϕ
3
2
10
3
⋅ 360
∘
ϕ
100
360
∘
= 2AU
1
6proton s
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.00secon d
(K . E . Moon)
(K . E . E ar th)
(Ear th Da y) ≈ 1secon d
G = 6.67408E − 11
m
3
kg ⋅ s
2
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
of 70 98
For a year we have
(365.25)(24)(60)(60)=31557600 seconds
And for an AU
1 AU=1.496E11m
We can immediately put this to work. In order for the earth
to stay in orbit its centripetal force must equal the
gravitational force. Its orbital velocity must be given by:
Since in our theory we present R as:
Then
G = 39.433
AU
3
M
⊙
⋅ year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
GM
⊙
= 40
AU
3
year
2
GM
⊙
4π
2
= 0.99885 ≈ 1
G
Mm
R
2
=
mv
2
r
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
of 71 98
Equation 8.3
Which evaluates:
Converting this to meters per second:
Which should be about right. The orbital velocity is given
in the data tables as about 30,000m/s is an average over a
varying velocity due to the Earth’s slightly elliptical orbit.
Since we have
For the earth, we show now it is true for the moon as well,
and indeed it would hold for any system circular, or
approximately circular. For the Sun and its planets
If you use astronomical units and Earth years. In general
we use
Which has the constant of proportionality
v =
5
3
GM
ϕ
v =
5
3
(40)(1)
0.618
= 6
AU
year
6
AU
year
⋅
year
3.156E 7
⋅
1.496E11m
AU
= 28,441
m
s
1
2
⋅
ϕ
100
360
∘
= 1AU
T
2
= a
3
T
2
= G
M + m
4π
2
a
3
1
k
= G
M + m
4π
2
of 72 98
If m is small compared with M, you can write
Since we want to work with Earth and the Moon m is not
so small compared to M. So,…
=
=
And these do round neatly to one-exponent like this. The
sidereal lunar month is
(27.83)(24)(60)(60)=2,404,513 seconds. Thus,…
The tables give
Which is an average for the lunar orbit in its slightly
elliptical orbit.
We need to find G in terms of Lunar Units (LU) as opposed
to Astronomical Units (AU), and in terms of the lunar
month as one instead of the earth year. We see G has the
same value we had for solar-earth units:
1
k
= G
M
4π
2
1
k
= (6.674E − 11)
(
5.9722E 24kg + 7.347731E 22kg
4π
2
)
(6.674E − 11)
(
6E 24kg
4π
2
)
4E14
4π
2
=
1E14
π
2
= 1E13
a
3
= (1E13)(2,404,512)
2
= 5.78E 25
a = 3.8668E8m
a = 3.833E8m
of 73 98
=
We only need to convert to m, kg, s by using their
relationships with LU, LM, and . That is, for the orbital
velocity is always six:
Where is the orbital velocity of the moon, and is the
orbital velocity of the earth. The difference comes when
converting the system from lunar units in the case of the
first to kg/m/s and from astronomical units in the case of
the second to kg/m/s. For example the first is:
Which is correct, the tables give that the lunar orbital
velocity is 1000 m/s. This factor of six is the factor
recurrent throughout our work in this paper, like
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(3.8668E8m)
3
⋅
5.9722E 24kg
M
⊕
⋅
(2,404,512s)
2
L M
2
40.896 ⋅
LU
3
M
⊕
⋅ L M
2
M
⊕
v
m
=
5
3
(40)(1)
ϕ
= 6
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
v
e
v
m
= 6
3.8668E8m
2,404,512s
= 964.886m /s
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
of 74 98
9.0 Logos
To find logos, is to find Nature not just as number,
but as dimensionless whole number, because
fractions have inherent in them irrational
numbers, which are unending decimal expressions
that at some point must be rounded to so many
figures to put them to use. Logos then is the
language of Nature that transcends conundrum
because it would be utilized as relationship
between point, plane, and line as opposed to units.
We have found that for the orbital velocity:
Equation 9.1
Where is the earth orbital radius. This results in the
orbital velocity for a orbiting body is
Equation 9.2.
Which is logos because it is number, which is 6. The orbital
velocity of a body is always 6 because:
Where is the orbital velocity of the moon, and is the
orbital velocity of the earth, That is G is always 40, and M
is always 1 providing the orbit is circular. Let us show this
for Venus. Its orbital distance VU (Venus units) is
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
e
v =
5
3
GM
ϕ
v
m
=
5
3
(40)(1)
ϕ
= 6
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
v
e
of 75 98
1.082E11meters. It orbital period is 1.94E7s is the Venus
year (VY).
=
Thus we have
We can then express all orbital velocities as 6, but to find
there values in a formal system of units, we need to convert
from these natural units to something like kg/m/s. Thus
logos translated into a language we understand can best be
done in a square array, or as a matrix transformation. We
have
You will find this gives ,
,
All of which are correct within the variations of these
velocities in their deviations from a perfectly circular orbit.
Logos is:
Equations 9.3
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
v
=
5
3
(40)(1)
ϕ
= 6
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
v
m
= 964.886m /s
v
e
= 28,443m /s
v
v
= 33,463.9m /s
of 76 98
, such that
which is given by
(
5
3
GM
ϕ
5
3
GM
ϕ
5
3
GM
ϕ
)
R
m
T
m
R
e
T
e
R
v
T
v
=
v
m
v
e
v
v
=
6
6
6
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1V U
v =
5
3
GM
ϕ
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
of 77 98
10.0 Conclusion To say that logos is:
Is to say that for the solar system Kepler’s Law of Planetary
Motion holds:
That in general
Because as such for Earth we have
For the Moon we have
For Venus we have
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
T
2
= a
3
T
2
= G
M + m
4π
2
a
3
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(3.8668E8m)
3
⋅
5.9722E 24kg
M
⊕
⋅
(2,404,512s)
2
L M
2
40.896 ⋅
LU
3
M
⊕
⋅ L M
2
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
V U
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
of 78 98
Making
, ,
This is derived from
Which is to say
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
=
5
3
(40)(1)
ϕ
= 6
v
v
=
5
3
(40)(1)
ϕ
= 6
G
Mm
R
2
=
mv
2
R
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1V U
v =
5
3
GM
ϕ
of 79 98
Is logos because
such that
which is given by
Which is the relationship between point, plane, and
line.
From everything we have said G, the gravitational
constant is about 40
We can do this for any planet and get G is approximately
40. We found the orbital velocity of any planets is 6. This
is true because as we have shown, the orbital distance of
any planet is
This gives since
That orbital velocities are for the moon, earth, and venus
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
AU
3
(Ear t h Distance Meters)
3
⋅
SolarMassKi logr ams
M
⊙
⋅
(Ear t hOrb i ta lPer io d Second s)
2
year
2
1
2
⋅
ϕ
100
360
∘
= 1AU
v =
GM
R
=
5
3
GM
ϕ
of 80 98
But what does this mean? It means since
And, v=6, that where the orbital velocities of the planets
are 6, their distances from the sun are all one.
This gives:
But where six is
Six is also
And one is
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
R =
GM
v
2
R =
40
36
≈ 1
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
1
1secon d
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton s = carbon(C )
of 81 98
And one is also
Where one second is given by Earth-moon orbital
mechanics
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
1
6secon d s
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton = hydr ogen(H )
(K . E . Moon)
(K . E . E ar th)
(Ear th Da y) ≈ 1secon d
of 82 98
11.0 Summary What we have done here is capture the
six-fold logos across the whole spectrum:
Which are the biological life skeletons the hydrocarbons.
Shown that carbon the core element of life is one second
predicts the radius of the proton:
That the second itself
Is a natural unit
We have brought this together:
Have described all orbital velocities as 6:
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d = hydrogen(H )
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
1
6
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
(K . E . Moon)
(K . E . E ar th)
(Ear th Da y) ≈ 1secon d
1
α
2
m
p
h 4π r
2
p
Gc
≈ 6proton s ⋅
K Eof Moon
K Eof Ear th
⋅ Ear th Day
of 83 98
Have brought in chemistries’ Avogadro’s number with a
constant k that that determines the integer 6 in terms of
the earth orbital velocity above:
,
But where six is
Six is also
And one is
And one is also
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
k v
e
= 6
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
1
1secon d
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton s = carbon(C )
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
of 84 98
Where one second is given by Earth-moon orbital
mechanics
1
6secon d s
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton = hydr ogen(H )
(K . E . Moon)
(K . E . E ar th)
(Ear th Da y) ≈ 1secon d
of 85 98
12.0 Graphic Equations We can say visually since the
orbital velocity of a planet or a moon is 6, that it is the
perimeter of a unit regular hexagon in logos:
Since the radius R of the orbit
in logos is 1, then its visual
logos is the radius R of this
regular hexagon because the
radius of a regular hexagon is
equal to its side (s). We have
,
,
The six equilateral triangles that make up the regular
hexagon are computed here:
The height of each equilateral triangle is . Since
then and since is a regular
hexagon, is a square because :
R =
1
2
⋅
ϕ
100
360
∘
v =
5
3
GM
ϕ
G = 40
3
2
3
2
= cos
π
6
3 = 2cos
π
6
π
6
π
4
π
4
= 45
∘
of 86 98
And, is a regular pentagon which contains is the
golden ratio is :
These are the regular polygons that tessellate (tile a plane
without leaving gaps). They tessellate in the plane, except
for the regular pentagon, which tessellates closed, it
encloses a volume, and is 1 of 4 other solids that do the
same which are made up of our tessellating regular
polygons, the equilateral triangle, the regular hexagon, and
the square:
π
5
Φ
1
ϕ
of 87 98
The area of a regular pentagon is given by
For s=1 this is . This gives the
surface are of a regular dodecagon is:
A =
1
4
5(5 + 2 5 s
2
A = 1.72 ≈ 1.732 = 3
of 88 98
Actually let us denote the surface area of the dodecagon as
And reserve A for the
area of a regular
pentagon, which we can
write either
Or,…
Much in the same way that in vector calculus
Are both surface integrals. We want to use surface area of a
pentagon as
∫
S
d x d y =
∬
S
d x d y
of 89 98
Even though it shows a dodecahedron, because it speaks of
the connection of the regular pentagon to the
dodecahedron. We want denote surface areas and volumes
V
= Volume of Dodecahedron
With actual images and shapes of these objects, because
logos is to have a deeper understanding of Nature, and this
allows us to work with the ideas in our mind in connection
to the equations. We have seen that the orbit of any planet
or moon can be taken as 1:
And that 1 is astronomical unit (AU), moon unit (MU),
Venus unit (VU), Mars unit (MU),…and so on. The
gravitational constant G can be taken as 40 for every
planet, but in units that correspond to the planet (or body)
in particular:
Where M is the mass of the body orbited, like the Sun for
the Earth, the Earth for the moon, and is the planet’s
orbital radius, it’s orbital period and these in any system
of units, for example in kg/m/s for the Earth
R
p
=
1
2
⋅
ϕ
100
360
∘
= 1
G = 6.674E − 11
m
3
kg
2
⋅ s
2
⋅
M
R
p
T
p
R
p
T
p
R
e
= 1.496E1m = 1AU = Ear thOrbit
of 90 98
For the moon is the orbital radius in meters from the
Earth, is the Earth mass, the mass of the body it orbits,
is its orbital period. The orbital velocity of any planet
can be taken as 6
And must be translated from this the logos to a system of
units we understand, like kg/m/s. In general the
components of logos are given by
n=1,2,3,4,5,6,…
In that
, , where ,
,…
We have orbital velocity is hexaperimeter, and orbital
radius is hexaradius:
M
⊙
= 1.98847E 30kg = SolarMa ss
T
e
= 31,557,600s = Ear thYear
R
m
M
⊕
T
m
v =
5
3
GM
ϕ
=
5
3
(40)(1)
0.618
= 6
f (n) = 2cos
π
n
2cos
π
6
= 3
2cos
π
5
= Φ
ϕ =
1
Φ
2cos
π
4
= 2
of 91 98
We are going to suggest that where orbital velocity is
associated with the regular hexagon, the orbital radius is
associated with the square, and the centripetal acceleration
is associated with the dodecahedron and pentagon. We
have the mass of the earth in solar masses is
Which is correct. Where
=
Thus we have for the orbit radii, orbital velocities, and
centripetal accelerations of an orbiting body
M
⊕
=
5.9722E 24
1.98847E 30
= 3.00E − 6M
⊙
= 0.00003M
⊙
a
e
=
v
2
e
R
e
=
6
2
1
= 36
AU
year
2
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
2cos(36
∘
) = 2cos
360
∘
10
=
1
2ϕ
360
∘
10
⋅
1
cos
−1
1
2ϕ
= 1
360
∘
10
= cos
−
1
1
2ϕ
= 36
∘
24 cos(36
∘
) ≈ 20
of 92 98
Thus we say
That,…
Is orbital radius, is:
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
(
6
2
6
2
6
2
)
3.8668E8m
(2,404,512s)
2
1.496E11m
(31,557,600s)
2
1.082E11m
(19,400,000s)
2
=
a
m
a
e
a
v
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
of 93 98
Is orbital velocity, is:
Is centripetal
acceleration, is:
Where we have used
The image of an icosahedron as the area
of a regular hexagon, because a regular
hexagon is its shadow.
Aside from our logos formulation of the planets being
useful theoretically, it is useful for orbital mechanics
computations whether it be of planets or spacecraft
because it uses multiplication of integer values like six for
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
(
6
2
6
2
6
2
)
3.8668E8m
(2,404,512s)
2
1.496E11m
(31,557,600s)
2
1.082E11m
(19,400,000s)
2
=
a
m
a
e
a
v
of 94 98
orbital velocity that can be converted to units we
understand in matrix form, which is at the heart of writing
computer programs in their linear algebra applications
which to the computer are arrays. Further since we
translate from logos to units, by dividing out large
numbers (like astronomical units and seconds in a year) in
such a way that they are not overly large numbers, yet
large enough that they can be written as integers without
losing much accuracy and there are many simple
programs on the internet that multiply matrices with
integer elements. This one comes from
programmingsimplified.com:!
#include <stdio.h>
int main(int argc, const char * argv[]) {
int m,n,p,q,c,d,k,sum=0;
int first[10][10],second[10][10],multiply[10]
[10];
printf("Enter number of rows and columns of
matrix 1: \n");
scanf("%d %d",&m,&n);
printf("Enter elements of matrix 1: \n");
for(c=0;c<m;c++)
for(d=0;d<n;d++)
scanf("%d",&first[c][d]);
printf("Enter number of rows and column of matrix
2: \n");
scanf("%d %d", &p,&q);
if(n!=p)
printf("The multiplication can't be done.");
else
{
printf("Enter the elements of matrix 2: \n");
for(c=0;c<p;c++)
for(d=0;d<q;d++)
scanf("%d", &second[c][d]);
for(c=0;c<m;c++){
for(d=0;d<q;d++){
for(k=0;k<p;k++){
sum=sum+first[c][k]*second[k][d];
}
of 95 98
multiply[c][d]=sum;
sum=0;
}
}
printf("Product of the matrices is: \n");
for(c=0;c<m;c++){
for(d=0;d<q;d++)
printf("%d\t", multiply[c][d]);
printf("\n");
}
}
return 0;
}
of 96 98
We need to divide out our data for the planets and the
moon, which are 9 elements, which works great because
our program works only up to 3 by 3 matrices. We have
{3.8668E8m}{2,404,512s^2}=161=moon
{1.496E11m}{31,557,600s^2}=4,740=earth
{1.082E11m}{19,400,000s^2}=5,577=venus
of 97 98
You can do the rest of the planets, but we will just do these
and put zeros in for the remaining 6 elements. Running the
program we have
Enter number of rows and columns of matrix 1:
3
3
Enter elements of matrix 1:
161
0
0
4740
0
0
5577
0
0
Enter number of rows and column of matrix 2:
3
3
Enter the elements of matrix 2:
6
0
0
6
0
0
6
0
0
Product of the matrices is:
966 0 0
28440 0 0
33462 0 0
These answers are in meters per second.These
answers are fairly accurate but can be made
more accurate by adjust the orbital velocity 6
according to orbital pecularities, which would
be a subject in itself that is outside the
scope of our work here.
of 98 98
The Author
